Two insulated circular loop A and B radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in…

Year	2026	2025	2024	Total
Questions	9	18	15	42

Q38 jee_main_2026_24_january_evening Galvanometer to Ammeter Conversion

A moving coil galvanometer of resistance 100 Ω shows a full scale deflection for a current of 1 mA. The value of resistance required to convert this galvanometer into an ammeter, showing full scale deflection for a current of 5 mA, is ____ Ω

A. $25$ $25$
B. $10$ $10$
C. $0.5$ $0.5$
D. $2.5$ $2.5$

Solution

Related Formula

rₛ = (ig \times G)/(i - ig) = (G)/(((i)/(ig) - 1))

Core Logic

Galvanometer to Ammeter Conversion diagram for Q38 - JEE Main 2026 Evening

Given: Galvanometer resistance, $G = 100 Ω$ $G = 100 \, \Omega$ Full scale deflection current, $ig = 1 mA$ $i_g = 1 \text{ mA}$ Target current, $i = 5 mA$ $i = 5 \text{ mA}$

Step 1: Calculate Shunt Resistance

Using the shunt resistance formula:

rₛ = (100)/(((5)/(1) - 1))

rₛ = (100)/(4) = 25 Ω

Pattern Recognition

When upgrading range $n$ $n$ times ( $n = i / ig$ $n = i / i_g$), the required parallel shunt is simply $G / (n - 1)$ $G / (n - 1)$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q34 jee_main_2026_28_january_morning Magnetic Force between Two Parallel Currents

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is ____.

Three parallel current carrying wires diagram for Q34 — Three vertical wires P, Q, R carrying 3A (up), 1A (down), and 2A (down) separated by 3cm and 2cm respectively.

(μ₀ = 4π \times 10⁻⁷T\cdot m / A)

A. $6 \times 10⁻⁷ ~N towards P$
B. $6 \times 10⁻⁶ ~N towards R$
C. $6 \times 10⁻⁷ ~N towards R$
D. $6 \times 10⁻⁶ ~N towards P$

Solution

Related Formula

F = (μ₀ I₁ I₂ )/(2π d)

Core Logic

Wires with current in opposite directions repel each other. Wires with current in the same direction attract each other. Calculate the individual force vectors on wire Q due to wires P and R.

Step 1: Force due to Wire P

Wire P (3A up) and Wire Q (1A down) carry current in opposite directions. So, P repels Q towards the right (towards R).

FPQ = (μ₀ IP IQ )/(2π d₁) = μ₀ (3)(1) 2π (3 \times 10⁻²)

Step 2: Force due to Wire R

Wire R (2A down) and Wire Q (1A down) carry current in the same direction. So, R attracts Q towards the right (towards R).

FRQ = (μ₀ IR IQ )/(2π d₂) = μ₀ (2)(1) 2π (2 \times 10⁻²)

Step 3: Net Force Calculation

Both forces act in the same direction (towards R).

Fₙₑₜ = FPQ + FRQ = (μ₀)/(2π) IQ ( (IP)/(d₁) + (IR)/(d₂) )

Fₙₑₜ = 2 \times 10⁻⁷ \times 1 \times ( 33 \times 10⁻² + 22 \times 10⁻² ) \times 15 \times 10⁻²

= 2 \times 10⁻⁷ \times ( 110⁻² + 110⁻² ) \times 15 \times 10⁻²

= 2 \times 10⁻⁷ \times (200) \times 0.15

= 400 \times 10⁻⁷ \times 0.15 = 60 \times 10⁻⁷ = 6 \times 10⁻⁶ ~N

Step 4: Direction

The net force is $6 \times 10⁻⁶ ~N$ $6 \times 10^{-6} \mathrm{~N}$ towards R.

Pattern Recognition

Always map the direction (attract/repel) before plugging in numbers. Same direction = Attract. Opposite direction = Repel.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q43 jee_main_2026_28_january_morning Magnetic Field on the Axis of a Circular Loop

The magnetic field at the centre of a current carrying circular loop of radius R is

16

. The magnetic field at a distance

x = \sqrt(3) R

on its axis from the centre is ____

.

A. $2\sqrt(2)$ $2\sqrt{2}$
B. $4$ $4$
C. $2$ $2$
D. $8$ $8$

Solution

Related Formula

Bcentre = (μ₀ I)/(2R)

Bₐₓᵢₛ = μ₀ I R²2(x² + R²)3/2

Core Logic

Substitute the expression for $Bcentre$ $B_{\text{centre}}$ into the equation for $Bₐₓᵢₛ$ $B_{\text{axis}}$ by evaluating the denominator with $x = \sqrt(3)R$ $x = \sqrt{3}R$.

Step 1: Using Center Field

Bcentre = (μ₀ I)/(2R) = 16

Step 2: Calculating Axial Field

Substitute $x = \sqrt(3)R$ $x = \sqrt{3}R$ into the axial field formula:

Bₐₓᵢₛ = μ₀ I R²2( (\sqrt(3)R)² + R² )3/2

= μ₀ I R²2(3R² + R²)3/2

= μ₀ I R²2(4R²)3/2

Step 3: Final Deduction

= (μ₀ I R²)/(2(8R³)) = (μ₀ I)/(16R)

= (1)/(8) ( (μ₀ I)/(2R) ) = (1)/(8) \times 16 = 2

Pattern Recognition

A classic standard derivation: at $x = \sqrt(3)R$ $x = \sqrt{3}R$, the field is $1/8$ $1/8$ of the center field. At $x = R$ $x = R$, it's $1/(2\sqrt(2))$ $1/(2\sqrt{2})$. Committing these to memory saves 60 seconds.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q39 jee_main_2026_28_january_evening Magnetic Field of a Cylindrical Conductor

A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is: A. maximum at either ends of the conductor and minimum at the midpoint B. maximum at the axis of the conductor C. minimum at the surface of the conductor D. minimum at the axis of the conductor E. same at all points in the cross-section of the conductor Choose the correct answer from the options given below:

A. $D Only$ $\text{D Only}$
B. $A, D Only$ $\text{A, D Only}$
C. $B, C Only$ $\text{B, C Only}$
D. $E Only$ $\text{E Only}$

Solution

Related Formula

For a solid cylinder carrying uniform current density: Inside ( $r < R$ $r < R$):

B = (μ₀ I r)/(2π R²)

Outside ( $r \geq R$ $r \geq R$):

B = (μ₀ I)/(2π r)

Core Logic

Solution for Magnetic Field of a Cylindrical Conductor

From the formula for the magnetic field inside the solid cylinder,

B

is directly proportional to the radial distance

r

from the axis. Thus, at the axis (

r = 0

),

B = 0

(which is its absolute minimum). At the surface (

r = R

),

B

reaches its maximum value

B = (μ₀ I)/(2π R)

.

Step 1: Evaluating the Statements

Statement A: Incorrect (B depends on radial distance, not longitudinal position). Statement B: Incorrect (B is minimum at the axis). Statement C: Incorrect (B is maximum at the surface). Statement D: Correct (B is zero at the axis). Statement E: Incorrect (B varies with $r$ $r$).

Pattern Recognition

Graph of B vs r for a uniform solid cylinder is a straight line passing through origin up to $r=R$ $r=R$, then a hyperbola $1/r$ $1/r$ outside. Zero at axis, max at surface.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Q2 jee_main_2025_02_april_evening Moving Coil Galvanometer

In a moving coil galvanometer, two moving coils

M₁

and

M₂

have the following particulars:

aligned R₁ &= 5 Ω, N₁ = 15, A₁ = 3.6 \times 10⁻³ m², B₁ = 0.25 T R₂ &= 7 Ω, N₂ = 21, A₂ = 1.8 \times 10⁻³ m², B₂ = 0.50 T aligned

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of

M₁

and

M₂

?

A. $1:1$ $1:1$
B. $1:4$ $1:4$
C. $1:3$ $1:3$
D. $1:2$ $1:2$

Solution

Related Formula

Voltage Sensitivity (Vₛ) = (θ)/(V) = (N B A)/(C R)

where: $N$ $N$ = number of turns $B$ $B$ = magnetic field $A$ $A$ = area of the coil $C$ $C$ = torsional constant of the spring $R$ $R$ = resistance of the coil

Core Logic

Since the torsional constant $C$ $C$ is the same for both coils, the ratio of voltage sensitivities of $M₁$ $M_1$ and $M₂$ $M_2$ is:

((Vₛ)₁)/((Vₛ)₂) = ((N₁ A₁ B₁)/(N₂ A₂ B₂)) \cdot ((R₂)/(R₁))

We are given the following values:

Coil 1: $R₁ = 5 Ω$ $R_1 = 5 \ \Omega$, $N₁ = 15$ $N_1 = 15$, $A₁ = 3.6 \times 10⁻³ m²$ , $B₁ = 0.25 T$ $B_1 = 0.25 \ \mathrm{T}$
Coil 2: $R₂ = 7 Ω$ $R_2 = 7 \ \Omega$, $N₂ = 21$ $N_2 = 21$, $A₂ = 1.8 \times 10⁻³ m²$ , $B₂ = 0.50 T$ $B_2 = 0.50 \ \mathrm{T}$

Step 1: Calculate the ratio

Substitute the values into the formula:

((Vₛ)₁)/((Vₛ)₂) = ( 15 \times 3.6 \times 10⁻³ \times 0.2521 \times 1.8 \times 10⁻³ \times 0.50) \times (7)/(5)

Simplify the terms within the brackets:

$3.6 \times 10⁻³1.8 \times 10⁻³ = 2$
$(0.25)/(0.50) = (1)/(2)$ $\frac{0.25}{0.50} = \frac{1}{2}$

(15 \times 2 \times (1)/(2))/(21) = (15)/(21) = (5)/(7)

Multiplying by $(R₂)/(R₁) = (7)/(5)$ $\frac{R_2}{R_1} = \frac{7}{5}$:

((Vₛ)₁)/((Vₛ)₂) = (5)/(7) \times (7)/(5) = 1

Thus, the ratio is $1:1$ $1:1$.

Pattern Recognition

Sees: Galvanometer sensitivity comparison with different parameters. Trap: Confusing Current Sensitivity with Voltage Sensitivity. Current sensitivity is $(NBA)/(C)$ $\frac{NBA}{C}$ (independent of $R$ $R$), while voltage sensitivity is $(NBA)/(C R)$ $\frac{NBA}{C R}$ (depends on $R$ $R$). Shortcut: Write the ratio as $((Iₛ)₁)/((Iₛ)₂) \times (R₂)/(R₁)$ to keep calculations clean.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Two insulated circular loop A and B radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be : Two circular current carrying loops placed orthogonally.

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Individual Fields

Step 2: Resultant Magnetic Field

Pattern Recognition

Chapter Mix

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate Shunt Resistance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Force due to Wire P

Step 2: Force due to Wire R

Step 3: Net Force Calculation

Step 4: Direction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Using Center Field

Step 2: Calculating Axial Field

Step 3: Final Deduction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating the Statements

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the ratio

Pattern Recognition

Chapter Mix

More Moving Charges and Magnetism Questions — jee_main_2024_30_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Two insulated circular loop A and B radius 'a' carrying a current of 'I' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :
Two circular current carrying loops placed orthogonally.